Let a^2 be an integer. Prove that a^2 is divisible by 5 if and only if a is divi

ID: 3036715 • Letter: L

Question

Let a^2 be an integer. Prove that a^2 is divisible by 5 if and only if a is divisible by 5. Prove that for m elementof N, if m > 2 and x = 1 mod m, y = 2 mod m, then xy = 2 mod m. Prove the following statement by considering three possible cases (values) of x mod 3: There is no integer x such that x^2 = 2 mod 3. Prove that, for all integers a and b, a^2 + b^2 = 0, 1, or 2 mod 4. Deduce that there do not exist integers a and b such that a^2 + b^2 = 1234567. For each n elementof Z, define P_n to be the set P_n = {17n, 17n + 1, 17n + 2..., 17n + 16}. Using the Division Lemma, prove that {P_n}_n elementof z is a partition of Z. (Recall the definition of a partition that appears as Definition 1.7 in the supplementary document. The Division lemma is necessary in the proof, so clearly indicate where you used the lemma in your proof.)

Explanation / Answer

1) As the question concerns mod 4, we may take both a and b to be any of 0,1,2 ,3.

Here is the table of values of (a2 +b2 ) (mod 4) for all possible combinations of a and b

Notice that (a2 +b2 ) (mod 4) is 0,1 or 2 mod 4 ,as required to be proved.

In particular (a2 +b2 ) (mod 4) is never 3 mod 4.

The number given 1234567 is 3 (mod 4), as it ends with 67.

So it cannot be expressed as a sum of squares a2 +b2

a 0 0 0 0 1 1 1 1 2 2 2 2 3 3 3 3 b 0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3 (a2 +b2 ) (mod 4) 0 1 0 1 1 2 1 2 0 1 0 1 1 2 1 2

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Let a^2 be an integer. Prove that a^2 is divisible by 5 if and only if a is divi

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